[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 376
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Это ж сколько лет этой шутке
лет 20 минимум
Is it the longer one or the green one?
P.S. Hint: it's longer. :)
Так больше длинный или зелёный?
long
on a cleverness test.
There is a number of rectangles of size 5x1. Put [one big rectangle without holes] together. Prove that the length of at least one of its sides is a multiple of 5.
Oh what a beauty!
By the way, you can replace the one in the condition with any other natural, and the original rectangles don't have to be all the same. Or even so:
There are some number of not necessarily identical rectangles, one side of which is a multiple of 5, and the other is equal to a natural number. One put together [one big rectangle without holes]. Prove that the length of at least one of its sides is a multiple of 5.
Mathemat писал(а) >>
There are a number of not necessarily identical rectangles, one side of which is a multiple of 5 and the other is equal to a natural number. Put [one big rectangle without holes] together. Prove that the length of at least one of its sides is a multiple of 5.
The area of the new rectangle
S=sum(5*x(i)*y(i))=X*Y;
5*sum(x(i)*y(i))=X*Y;
since x(i) y(i) is a natural number
then X or Y is divisible by 5;
20 chess players played a one-round tournament (each played one game against each other). A correspondent of "Sportivnaya Gazeta" wrote in his article that each participant of this tournament won as many games as he drew. Prove that the correspondent was wrong.
1 chess player played with 19 others, and 19 is not divided by 2 without a remainder.
1 шахматист играл с 19-ю другими, а 19 на 2 без остатка не делится.
Why two?
Who did they beat?
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